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I've read through some of the questions on line-plane intersection but I'm struggling because many define the plane in a format like:

$4x-2y+5z-9=0$,

but my plane is defined as a point ($x_p, y_p, z_p$) and the unit vector normal to the plane ($i_p, j_p, k_p$).

My line is defined by a point ($x_v, y_v, z_v$) and a unit vector in the direction of the line ($i_v, j_v, k_v$),

I can't figure out how to apply what I've read in other questions to my scenario, which I will need to implement in some generic VB.NET code.

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  • $\begingroup$ Welcome to Mathematics Stack Exchange. It might be easier to explain with a numerical example, but basically the equation of your plane is $i_px+j_py+k_px-(i_px_p+j_py_p+k_pz_p)=0$; on the other hand, the normal vector to the plane $4x-2y+5z-9=0$ is $(4,-2,5)$ $\endgroup$ – J. W. Tanner Oct 28 '19 at 10:20
  • $\begingroup$ You’re only one simple step away from an equation of that form for the plane. Look up “point-normal form.” $\endgroup$ – amd Oct 28 '19 at 17:03
  • $\begingroup$ @amd only simple if you know. Fortunately AugSB’s answer got me there. $\endgroup$ – Notts90 supports Monica Oct 28 '19 at 17:04
  • $\begingroup$ You could’ve gotten there yourself. I’d bet that he started with that form, substituted a generic point on the line into it and then solved for the unknown multiple of the line’s unit direction vector. $\endgroup$ – amd Oct 28 '19 at 17:06
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Let $P_v=(x_p,y_p,z_p)$, $P_p=(x_v,y_v,z_v)$, $\vec{v_v}=(i_v,j_v,k_v)$ and $\vec{v_p}=(i_p,j_p,k_p)$. Then

$$P_{intersection} = P_v + \frac{(P_p-P_v)\cdot\vec{v_p}}{\vec{v_v}\cdot\vec{v_p}} \ \vec{v_v}$$

I assume you already know the basics of operations with vectors and points...


EDIT: This formula can be easily derived from the vectorial definition of planes and lines. A point $P$ belongs to the plane defined by $P_p$ and $\vec{v_p}$ if $(P-P_p)\cdot\vec{v_p}=0$. Moreover, points of the line defined by $P_v$ and $\vec{v_v}$ are of the form $P_v+\lambda \vec{v_v}$. Note that the intersection point has to satisfy both conditions, so it is enouh to plug in the line form into the plane equation and solve: $$(P_v+\lambda \vec{v_v}-P_p)\cdot\vec{v_p}=0 \iff \lambda =\frac{(P_p-P_v)\cdot\vec{v_p}}{\vec{v_v}\cdot\vec{v_p}}$$ Of course, if $\vec{v_v}\cdot\vec{v_p}=0$, both elements would be parallel, so there would not be any intersection point.

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  • $\begingroup$ Its been a while but yes! Use dot product on the vectors right? I best get refreshing... $\endgroup$ – Notts90 supports Monica Oct 28 '19 at 10:31
  • $\begingroup$ This answer would be much improved if you explained how you arrived at this mysterious formula. $\endgroup$ – amd Oct 28 '19 at 17:07
  • $\begingroup$ @amd mysterious answer that works great. $\endgroup$ – Notts90 supports Monica Oct 28 '19 at 17:15
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    $\begingroup$ @amd It is not so misterious, you can find it in plenty of webpages (starting from Wikipedia itself). Anyhow, I added a brief explanation to complete the answer. $\endgroup$ – AugSB Oct 29 '19 at 7:47
  • $\begingroup$ It is mysterious when you present it as a fact accompli with no explanation or references. $\endgroup$ – amd Oct 29 '19 at 16:24
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If $\vec P=(x,y,z)$ is some point of the plane that is given by point $\vec A$ and a normal $\vec n$, then by definition of normal the vector $\vec{AP}=\vec P -\vec A$ is perpendicular to $\vec n$, so the scalar product $\vec{AP}\cdot\vec n=0$.

By writing this in coordinate form, you will get the plane equation: $$ (x-x_v)i_v+(y-y_v)j_v+(z-z_v)k_v = 0,\\ i_vx+j_vy+k_vz-(i_vx_v+j_vy_v+k_vz_v)=0 $$

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